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Question

An engine expends 40.0hp40.0 \mathrm{hp} in moving a car along a level track at a speed of 15.0 m/s15.0 \mathrm{~m} / \mathrm{s}. How large is the total force acting on the car in the direction opposite to the motion of the car?

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Answered 1 year ago
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Given the following data

P=40.0 hp(746 W1 hp)=29840 Wv=15.0 ms\begin{aligned} P&=40.0\ \mathrm{hp}\left(\dfrac{746\ \mathrm{W}}{1\ \mathrm{hp}}\right)\\ &=29840\ \mathrm{W}\\ v&=15.0\ \mathrm{\dfrac{m}{s}}\\ \end{aligned}

our objective is to calculate the total force FF exerted on the car in the direction opposite to that of the motion.

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